Transcript PowerPoint - Limiting Reagents Lab

1) 2) CaCl2•2H2O(aq) + Na2CO3(aq)
 2H2O + CaCO3(s) + 2NaCl(aq)
3) # mol CaCl2•2H2O=
2.00 g CaCl2•2H2Ox 1 mol CaCl2•2H2O = 0.0136 mol
147.02 g CaCl2•2H2O CaCl2•2H2O
# mol Na2CO3 =
1.00 g Na2CO3 x 1 mol Na2CO3 = 0.00943 mol
Na2CO3
105.99 g Na2CO3
What we have
What we need
Na2CO3
CaCl2•2H2O
0.0136 mol 0.00943 mol
.0136/.00943 .00943/.00943
= 1.44 mol = 1 Limiting
1 mol
1 mol
4) CaCl2•2H2O is in excess. It is aqueous, so it
will stay dissolved in water, ending up in the
filtrate. Adding Na2CO3 to the filtrate will make
it cloudy, revealing the presence of CaCl2.
5) # g NaCl =
0.00943 mol Na2CO3 x 2 mol NaCl x 58.44 g NaCl
1 mol Na2CO3 1 mol NaCl
= 1.1027 g NaCl
or
1.00 g Na2CO3 x
1 mol Na2CO3
105.99 g Na2CO3
6) # g CaCO3 =
0.00943 mol Na2CO3 x 1 mol CaCO3 x100.09 g CaCO3
1 mol Na2CO3 1 mol CaCO3
= 0.944 g CaCO3
7) # g CaCO3 =
0.0136 mol CaCl2•2H2O x
1 mol CaCO3 100.09 g CaCO3
x
1 mol CaCl2•2H2O 1 mol CaCO3
= 1.36 g CaCO3
8) Hopefully the prediction was correct (closer
to 6). The comparison of 6 vs. 7 gives us a
shortcut for determining limiting reagents.
6) # g CaCO3 =
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0.00943 mol Na2CO3 x 1 mol CaCO3 x100.09 g CaCO3
1 mol Na2CO3 1 mol CaCO3
or
= 0.944 g CaCO3
1.00 g Na2CO3 x 1 mol Na2CO3
105.99 g Na2CO3
7) # g CaCO3 =
1 mol CaCO3 100.09 g CaCO3
x
1 mol CaCl2•2H2O 1 mol CaCO3
or
= 1.36 g CaCO3
2.00 g CaCl2•2H2O x 1 mol CaCl2•2H2O
147.02 g CaCl2•2H2O
0.0136 mol CaCl2•2H2O x